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JEE Main 20246 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Limits

JEE Main 2024 · 6 April · Evening Shift · Mathematics · Limits

_ n arrow ∞ ((1^2-1)(n-1)+(2^2-2)(n-2)+ +((n-1)^2-(n-1)) ·...

lim⁡n→∞(12−1)(n−1)+(22−2)(n−2)+⋯+((n−1)2−(n−1))⋅1(13+23+⋯⋯+n3)−(12+22+⋯⋯+n2)\lim _{n \rightarrow \infty} \frac{\left(1^2-1\right)(n-1)+\left(2^2-2\right)(n-2)+\cdots+\left((n-1)^2-(n-1)\right) \cdot 1}{\left(1^3+2^3+\cdots \cdots+n^3\right)-\left(1^2+2^2+\cdots \cdots+n^2\right)} is equal to :

Options

  1. A23\frac{2}{3}
  2. B13\frac{1}{3}
  3. C34\frac{3}{4}
  4. D12\frac{1}{2}

Correct answer

B. 13\frac{1}{3}

Step-by-step solution

lim⁡n→∞∑r=1n−1(r2−r)(n−r)∑r=1nr3−∑r=1nr2lim⁡n→∞∑r=1n−1(−r3+r2(n+1)−nr)(n(n+1)2)2−n(n+1)(2n+1)6lim⁡n→∞(((n−1)n)2)2+(n+1)(n−1)n(2n−1)6−n2(n−1)2n(n+1)2(n(n+1)2−2n+13)lim⁡n→∞n(n−1)2(−n(n−1)2+(n+1)(2n−1)3−n)n(n+1)23n2+3n−4n−26lim⁡n→∞(n−1)(−3n2+3n+2(2n2+n−1)−6)(n+1)(3n2−n−2)lim⁡n→∞(n−1)(n2+5n−8)(n+1)(3n2−n−2)=13\begin{aligned} & \lim _{n \rightarrow \infty} \frac{\sum_{r=1}^{n-1}\left(r^2-r\right)(n-r)}{\sum_{r=1}^n r^3-\sum_{r=1}^n r^2} \\ & \lim _{\mathrm{n} \rightarrow \infty} \frac{\sum_{\mathrm{r}=1}^{\mathrm{n}-1}\left(-\mathrm{r}^3+\mathrm{r}^2(\mathrm{n}+1)-\mathrm{nr}\right)}{\left(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)^2-\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}} \\ & \lim _{n \rightarrow \infty} \frac{\left(\frac{((n-1) n)}{2}\right)^2+\frac{(n+1)(n-1) n(2 n-1)}{6}-\frac{n^2(n-1)}{2}}{\frac{n(n+1)}{2}\left(\frac{n(n+1)}{2}-\frac{2 n+1}{3}\right)} \\ & \lim _{n \rightarrow \infty} \frac{\frac{n(n-1)}{2}\left(\frac{-n(n-1)}{2}+\frac{(n+1)(2 n-1)}{3}-n\right)}{\frac{n(n+1)}{2} \frac{3 n^2+3 n-4 n-2}{6}} \\ & \lim _{n \rightarrow \infty} \frac{(n-1)\left(-3 n^2+3 n+2\left(2 n^2+n-1\right)-6\right)}{(n+1)\left(3 n^2-n-2\right)} \\ & \lim _{n \rightarrow \infty} \frac{(n-1)\left(n^2+5 n-8\right)}{(n+1)\left(3 n^2-n-2\right)}=\frac{1}{3} \\ & \end{aligned}
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